我试图用投影仪制作一个图形，显示一个函数，它的导数和指向该函数的法向量，但我无法编译，它说
Runaway argument?
\par \draw [line width=1.2pt,color=blue!50!black,smooth,samples=100,\ETC.
! File ended while scanning use of \frame.
<inserted text>
\par
我知道这些错误与留下一些不好的匹配（没有结束的开始）有关，但我没有发现任何类似的东西，并且绘制命令中的选项对我来说似乎并不奇怪。
MWE（应该是）是：
\documentclass[spanish]{beamer}
\pgfplotsset{compat=1.17}
\usepackage{soul}
\usepackage[T1]{fontenc}
\usepackage{amsmath}
\usepackage{tikz, xcolor}
\usetikzlibrary{shapes,arrows}
\usetikzlibrary{decorations.pathreplacing}
\usetikzlibrary{patterns,hobby}
\begin{document}
\begin{frame}
\frametitle{Aplicaciones geométricas de la derivada}
Sea una curva $C$ de ecuación $y=f(x)$ donde $f(x)$es continua
La pendiente de la recta tangente a la curva $C$en el punto $P(x_{0},y_{0}$)
es $m=f\prime(x_{0})$.
Usando la ecuación punto-pendiente de la recta:
\begin{center}
$y-y_{0}=m(x-x_{0})$
\end{center}
\begin{center}
$y-y_{0}=f\prime(x_{0})(x-x_{0})$
\end{center}
donde $x_{0}$ y $y_{0}$ son las coordenadas del punto de tangencia
\begin{tikzpicture}[scale=1]
\draw (-1,0) -- (7.5,0) (0,-.5) -- (0,4.5);
\node[below left=2pt and 2pt] at (0,0) {$O$};
\begin{scope}
\clip (-1,0) rectangle (7.5,4.5);
\draw[line width=1.2pt,color=red,smooth,samples=100,domain=-2.5:7.5] plot(\x,{0.3*((\x)-3.5)*((\x)-3.5)+0.5});
\end{scope}
\def\xA{2.5} \def\yA{0.8}
\coordinate (A) at (\xA,\yA);
\draw[dashed] (\xA,0) node[below=2pt] {$x_{0}$} -- (A) -- (0,\yA) node[left] {$y_{0}$};
\def\xM{6.5} \def\yM{3.2}
\coordinate (M) at (\xM,\yM);
\draw[dashed] (\xM,0) node[below] {$x_{0}+\Delta x$} -- (M) -- (0,\yM) node[left] {$y_{0}+\Delta x)$};
\begin{scope}
\clip (0.5,-0.5) rectangle (7.5,4.5);
\foreach \xN/\yN in {5}
{
\coordinate (N) at (\xN,\yN);
\tkzDrawPoint[size=8](N)
\tkzDrawLine[add=2 and 3,color=orange!50!black](A,N)
\tkzLabelPoint[line width=1.2pt,color=green!50!black,smooth,samples=100,domain=0.5:5](3,3){Rectas secantes}
}
\end{scope}
\draw[line width=1.2pt,color=blue!50!black,smooth,samples=100,domain=0.5:5] plot(\x,{-0.6*(\x)+2.3}) node[below] {Recta tangente a la curva en $P$};
\tkzDrawPoints[size=8](A,M)
\tkzLabelPoint[above=3pt](A){$P$}
\tkzLabelPoint[above left](M){$Q$}
\end{tikzpicture}
\end{frame}
\end{document}
当它在没有框架环境的情况下编译时，它会得到下一张幻灯片 
更新
定义 \pgfplotsset 和适当的库之后，MWE
\documentclass[spanish]{beamer}
\usepackage[T1]{fontenc}
\usepackage[latin9]{luainputenc}
\usepackage{xcolor}
\usepackage{calc}
\usepackage{amsmath}
\PassOptionsToPackage{normalem}{ulem}
\usepackage{ulem}
\usepackage{subcaption}
\usepackage{pifont}
\usepackage{pgfplots}
\usepackage{tkz-euclide}
\usetikzlibrary{angles, quotes}
\usepackage{amsmath}
\usepackage{setspace}
\usepackage{amsthm}
\usetikzlibrary{decorations.markings}
\usepackage[makeroom]{cancel}
\usetikzlibrary{datavisualization.formats.functions}
\usepackage{pgfplotstable}
\usepackage{wasysym}
\usetikzlibrary{decorations.pathreplacing}
\setbeamertemplate{navigation symbols}{}
\pgfplotsset{compat=1.17}
\usepackage{soul}
\usepackage{amssymb}
\usepackage{enumitem}
\usepackage{pifont}
\usetikzlibrary{3d}
\usepackage{tikz, xcolor}
\usetikzlibrary{shapes,arrows}
\usetikzlibrary{decorations.pathreplacing}
\usetikzlibrary{patterns,hobby}
\begin{document}
\begin{frame}
\frametitle{Aplicaciones geométricas de la derivada}
Sea una curva $C$ de ecuación $y=f(x)$ donde $f(x)$es continua
La pendiente de la recta tangente a la curva $C$en el punto $P(x_{0},y_{0}$)
es $m=f\prime(x_{0})$.
Usando la ecuación punto-pendiente de la recta:
\begin{center}
$y-y_{0}=m(x-x_{0})$
\end{center}
\begin{center}
$y-y_{0}=f\prime(x_{0})(x-x_{0})$
\end{center}
donde $x_{0}$ y $y_{0}$ son las coordenadas del punto de tangencia
\begin{tikzpicture}[scale=1]
\draw (-1,0) -- (7.5,0) (0,-.5) -- (0,4.5); % Axis
\node[below left=2pt and 2pt] at (0,0) {$O$}; % Origin
\begin{scope}
\clip (-1,0) rectangle (7.5,4.5);
\draw[line width=1.2pt,color=red,smooth,samples=100,domain=-2.5:7.5] plot(\x,{0.3*((\x)-3.5)*((\x)-3.5)+0.5});
\end{scope}
\def\xA{2.5} \def\yA{0.8}
\coordinate (A) at (\xA,\yA);
\draw[dashed] (\xA,0) node[below=2pt] {$x_{0}$} -- (A) -- (0,\yA) node[left] {$y_{0}$};
\def\xM{6.5} \def\yM{3.2}
\coordinate (M) at (\xM,\yM);
\draw[dashed] (\xM,0) node[below] {$x_{0}+\Delta x$} -- (M) -- (0,\yM) node[left] {$y_{0}+\Delta x)$};
\begin{scope}
\clip (0.5,-0.5) rectangle (7.5,4.5);
\foreach \xN/\yN in {5}
{
\coordinate (N) at (\xN,\yN);
\tkzDrawPoint[size=8](N)
\tkzDrawLine[add=2 and 3,color=orange!50!black](A,N)
\tkzLabelPoint[line width=1.2pt,color=green!50!black,smooth,samples=100,domain=0.5:5](3,3){Rectas secantes}
}
\end{scope}
\draw[line width=1.2pt,color=blue!50!black,smooth,samples=100,domain=0.5:5] plot(\x,{-0.6*(\x)+2.3}) node[below] {Recta tangente a la curva en $P$};
\tkzDrawPoints[size=8](A,M)
\tkzLabelPoint[above=3pt](A){$P$}
\tkzLabelPoint[above left](M){$Q$}
\end{tikzpicture}
\end{frame}
\end{document}
答案1
编辑（1）： 为了提供解决方案，在附录框架中添加了给定函数 f(x) 的切线计算程序。
编辑（2）：
似乎提供的结果图像有误：割线应该经过点 P 和 Q（即
变为切线）。通过对 MWE 的切线计算表示进行这种修正，切线计算的表示变得正确，并且其代码也变得更简单一些。
图表的格式略有修改。由于只使用了pfplots包，因此正切是通过函数导数计算的（参见 MWE 中的附录），对于方程组，定义计算正切的规则使用包align中定义的环境amsmath（由 加载beamer）：
\documentclass[spanish]{beamer}
\setbeamertemplate{frametitle continuation}{\hfill%
(\insertcontinuationcount)}
\usepackage[T1]{fontenc}
\usepackage{pgfplots}
\pgfplotsset{compat=1.17}
\usepackage[low-sup]{subdepth}
\begin{document}
\begin{frame}[fragile]
\frametitle{Aplicaciones geométricas de la derivada}
Sea una curva $C$ de ecuación $y=f(x)$ donde $f(x)$ es continua
La pendiente de la recta tangente a la curva $C$en el punto $P(x_{0},y_{0})$ es $m=f'(x_{0})$. Usando la ecuación punto-pendiente de la recta:
\begin{align*}
y-y_{0} & = m(x-x_{0}) \\
y-y_{0} & = f'(x_{0})(x-x_{0})
\end{align*}
donde $x_{0}$ y $y_{0}$ son las coordenadas del punto de tangencia:
\begin{center}
\begin{tikzpicture}[
dot/.style = {circle, fill, inner sep=2pt,
fill opacity=0.5, text opacity=1,
label={[inner sep=1pt]#1}, node contents={}},
every node/.append style = {sloped, align=center,
font=\scriptsize, text depth=0.25ex},
]
\begin{axis}[width=90mm, height=60mm,
axis lines=middle,
ylabel=y,
ylabel style = {rotate=270},
xlabel=x,
ymin=-1.1, ymax=5.0, ytick=\empty,
xmin=-1.5, xmax=7.5, xtick=\empty,
no marks
]
\node[below left] at (0,0) {0}; % Origin
%
\addplot +[ultra thick, samples=101,domain=-0.2:7.2]
plot {0.3*(\x-3.5)^2 + 0.5};
% tangent
\draw[red, shorten <=-2mm, shorten >=-9mm ]
(0,2.975) -- (3.3,0)
node[at end] {Recta tangente\\ a la curva en $P$};
% P and Q coordinates
\draw[densely dashed]
( 2.0,-0.10) node[below] {$x_0$} |-
(-0.1,1.175) node[left] {$y_0$}
node (p) [pos=0.5, dot=$P$];
\draw[densely dashed]
(6.5,-0.1) node[below] {$x_0+\Delta x$} |-
(-0.1,3.2) node[left] {$y_0+\Delta y$}
node (q) [pos=0.5, dot=$Q$];
% secant
\draw[purple, shorten <=-12mm, shorten >=-9mm]
(p) -- node {Rectas\\ secantes} (q);
\end{axis}
\end{tikzpicture}
\end{center}
\end{frame}
\begin{frame}[fragile, allowframebreaks]
\frametitle{Appendix:\\
calculation of the $f(x)$ tangent at the point $P(x_0,y_0)$}
\small
Rule (again):
{\color{blue}
\begin{align*}
y-y_{0} & = m(x-x_{0}) \\
y-y_{0} & = f'(x_{0})(x-x_{0})
\end{align*}
}
Graph $p(x)$ at point $P(x_0=2,y)$:
\begin{align*}
f(x) & = 0.3(x-3.5)^2 + 0.5 \\
P(2) & = 0.3(2-3.5)^2 + 0.5 = 0.3(-1.5)^2 + 0.5 = \boxed{1.175}
\end{align*}
\begin{flalign*}
\text{Point $P$:}
&& P(2,y) & = P\bigl(2,p(2)\bigr) = (2,0.3(2-3.5)+0.5) && \\
&& & = \boxed{$P(2,1.175)$} &&
\end{flalign*}
Derivative of $f(x)$:
\begin{align*}
f'(x) & = 0.3.2(x-3.5)^1 = 0.6x - 2.1 \\
f'(2) & = 1.2 - 2.1 = \boxed{-0.9}
\end{align*}
\framebreak
Tangent:
\begin{align*}
y - 1.175
& = f'(2)(x - 2) \Rightarrow y = -0.9(x - 2) = -0.9x +1.8 \\
y & = -0.9x + 1.8 + 1.175 \\
& = \boxed{-0.9x + 2.75}
\end{align*}
\begin{align*}
\text{value $y$ at $x=0$:} && y & = 2.975 \\
\text{value $x$ at $y=0$:} && x & = \frac{2.975}{0.9} = 3.305^\cdot
\end{align*}
\begin{center}
\begin{tikzpicture}[
dot/.style = {circle, fill, inner sep=2pt,
fill opacity=0.5, text opacity=1,
label={[inner sep=1pt]#1}, node contents={}},
every node/.append style = {sloped, font=\scriptsize, align=center},
every pin/.style = {pin edge={<-}, inner sep=1pt,
pin distance=3mm, font=\tiny}
]
\begin{axis}[width=90mm, height=45mm,
axis lines=middle,
ylabel=y,
ylabel style = {rotate=270},
xlabel=x,
ymin=-1.5, ymax=5.5, ytick=\empty,
xmin=-2.0, xmax=7.5, xtick=\empty,
no marks
]
\node[below left] at (0,0) {0}; % Origin
\addplot +[ultra thick, samples=101,domain=-0.2:7.2]
plot {0.3*(\x-3.5)^2 + 0.5};
% tangent
\draw[red, shorten <=-2mm, shorten >=-9mm ]
(0,2.975) coordinate (a) --
(3.3,0) coordinate (b)
node[at end] {Recta tangente\\ a la curva en $P$};
\path (a) node[inner sep=0pt, pin=225:2.975]{}
(b) node[inner sep=0pt, pin=265:$3.305^\cdot$] {};
% P and Q coordinates
\draw[densely dashed]
( 2.0,-0.10) node[below] {$x_0=2$} |-
(-0.1,1.175) node[left] {$y_0$=1.175}
node (p) [pos=0.5, dot=$P$];
\draw[densely dashed]
(6.5,-0.1) node[below] {$x_0+\Delta x$} |-
(-0.1,3.2) node[left] {$y_0+\Delta y$}
node (q) [pos=0.5, dot=$Q$];
% secant
\draw[purple, shorten <=-12mm, shorten >=-9mm]
(p) -- node {Rectas\\ secantes} (q);
\end{axis}
\end{tikzpicture}
\end{center}
\end{frame}
\end{document}
（未显示带附录的框架）